Potential at a General Point Due to a Dipole
Trending Questions
- (pEI)1/2
- (pEI)3/2
- (IpE)1/2
- (pIE)1/2
- 8×10−5 Nm
- 8√2×10−5 Nm
- 8√2×10−5 Nm
- 2√2×10−5 Nm
- 15 C-m along AB
- 20 C-m along AC
- 25 C-m along BC
- 25 C-m along CB
- K pcos2ϕr2
- zero
- 2K pcos2ϕr2
- K pcos ϕr2
(Take V=0 at infinity)
- 0, |→p|4πε0d2
- 0, −→p4πε0d3
- |→p|4πε0d2, →p4πε0d3
- 0, →p4πε0d3
- −20×10−18J
- −7×10−27J
- −10×10−29J
- −9×10−20J
- −1 Joule
- 1 Joule
- 2 Joules
- 5 Joules
A 4 μF capacitor is given 20 μC charge and is connected with an uncharged capacitor of capacitance 2 μF as shown in figure. When switch S is closed –
Charged flown through the battery is 403 μC
Charge flown through the battery is 203 μC
Work done by the battery is 2003 μJ
Work done by the battery is 1003 μJ
- p4πϵ0r2
- 0
- p2πϵ0r2
- p8πϵ0r2
- 0.1×10−4 Nm
- 1.13×10−4 Nm
- 2.5×10−4 Nm
- 3.7×10−4 Nm
- (60 ^i) N
- (−60 ^i) N
- (30 ^i) N
- (−30 ^i) N
- Potential energy of dipole is greater for orientations (1) and (3) than (2) and (4).
- If the dipole rotates from orientation (1) to (2), then the workdone on the dipole by field is positive.
- If the dipole rotates from orientation (1) to (2), then the workdone on the dipole by field is negative.
- Workdone to rotate dipole from (1) to (2) is same as workdone to rotate dipole from (1) and (4).
- The total work done on the charge is zero.
- The work done by the electrostatic force from A to C is negative.
- The work done by the electrostatic force from C to B is positive.
- The work done by electrostatic force in taking the charge from A to B is dependent on the actual path.
- θ=90∘
- θ=180∘
- θ=30∘
- θ=0∘
- 15 C-m along AB
- 20 C-m along AC
- 25 C-m along BC
- 25 C-m along CB
- V=p cos θ4πϵ0r2
- V=p cos θ4πϵ0r
- V=p sin θ4πϵ0r
- V=p cos θ2πϵ0r2
A charge +Q is placed at the centre. The work done by external agent in taking the charge from centre to infinity is
- Q28πε0a−Q28πε0b
- −Q24πε0a−Q24πε0b+Q24πε0(a−b)
- −Q28πε0b
- Q28πε0a
- 1r
- 1r4
- 1r3
- 1r2
- the electric field →E at all points on the x− axis has same direction
- →E at all points on the y− axis is along ^i
- →E at all points on the y− axis is along −^i
- →E at all points on the y− axis is along ^j
Current I is flowing through two materials having electrical conductivities σ1 and σ2 where σ1>σ2, then total amount of charge at the junction of materials is
- zero
- positive
- negative
- may be positive or negative
- (+^i−3^j−2^k)
- (−^i+3^j−2^k)
- (+^i+3^j−2^k)
- (−^i−3^j+2^k)
- V=p cosθ4πϵ0r2
- V=p cosθ4πϵ0r
- V=p sinθ4πϵ0r
- V=p cosθ2πϵ0r2
- 2π√IpE
- 12π√pEI
- π√IpE
- 2π√IEp
[0.77 Mark]
- The direction of electrostatic field will be perpendicular to the equipotential surface
- The potential difference between any two points on an equipotential surface is zero
- For a point charge, the equipotential surfaces will be concentric spheres
- All the statements above are correct.
- EinsideA=0
- QA>QB
- σAσB=RBRA
- EonsurfaceA<EonsurfaceB
- (pEI)12
- (pEI)32
- (IpE)12
- (pIE)12
(C is centre of dipole of moment P2)
- 2k P1P2cosθr3
- −2kP1P2cosθr3
- −2k P1P2sinθr3
- −4k P1P2cosθr3
This section contains 1 Matrix Match type question, which has 2 Columns (Column I and Column II). Column I has four entries (A), (B), (C) and (D), Column II has four entries (P), (Q), (R) and (S). Match the entries in Column I with the entries in Column II. Each entry in Column I may match with one or more entries in Column II.
इस खण्ड में 1 मैट्रिक्स मिलान प्रकार का प्रश्न है, जिसमें 2 कॉलम (कॉलम I तथा कॉलम II) हैं। कॉलम I में चार प्रविष्टियाँ (A), (B), (C) तथा (D) हैं, कॉलम II में चार प्रविष्टियाँ (P), (Q), (R) तथा (S) हैं। कॉलम I में दी गयी प्रविष्टियों का मिलान कॉलम II में दी गयी प्रविष्टियों के साथ कीजिए। कॉलम I में दी गयी प्रत्येक प्रविष्टि का मिलान कॉलम II में दी गयी एक या अधिक प्रविष्टियों के साथ हो सकता है।
Match the Columns and choose the appropriate answer
कॉलमों का मिलान करें और सही उत्तर चुनें
Consider the following electrostatic field lines due to a positive point charge. Four points A, B, C & D are also shown in the figure.एक धनात्मक बिन्दु आवेश के कारण नीचे दी गई स्थिरवैद्युत क्षेत्र रेखाओं पर विचार कीजिए। चित्र में चार बिन्दु A, B, C व D भी दर्शाए गए हैं।
Match the entries given in Column-I with information given in Column-II.
कॉलम-I में दी गई प्रविष्टियों का मिलान कॉलम-II में दी गई सूचना के साथ कीजिए।
Column I कॉलम I |
Column II कॉलम II |
||
(A) | At A, as compared to B B की तुलना में A पर |
(P) | Potential is high विभव उच्च है |
(B) | At C, as compared to A A की तुलना में C पर |
(Q) | Potential is low विभव निम्न है |
(C) | At B, as compared to C C की तुलना में B पर |
(R) | Magnitude of electric field is low विद्युत क्षेत्र का परिमाण निम्न है |
(D) | At D, as compared to B B की तुलना में D पर |
(S) | Magnitude of electric field is high विद्युत क्षेत्र का परिमाण उच्च है |
- A-PS; B-QR; C-PS; D-QR
- A-P; B-Q; C-PS; D-QR
- A-S; B-Q; C-PS; D-R
- A-PQS; B-PQR; C-PS; D-QRS